Question

已知y+=(x+6)的6次方分之根号下x-3✖️(x+2)的2次方,则dx+分之dy

Answer 1

y = √[(x-3) (x+2)^2[/(x+6)^6,
y'=(1/2)[(x+2)^2+2(x-2)(x+2)]^(-1/2)/(x+6)^6
-6 √[(x-3) (x+2)^2]/(x+6)^7.

Answer 2

y = √(x-3) (x+2)^2/(x+6)^6
lny = (1/2)ln(x-3) + 2ln(x+2) - 6ln(x+6)
y'/y = (1/2)/(x-3) + 2/(x+2) - 6/(x+6)
dy/dx = y' = [√(x-3) (x+2)^2/(x+6)^6][(1/2)/(x-3) + 2/(x+2) - 6/(x+6)]